[Leetcode] Interleaving String

最新推荐文章于 2019-09-18 20:15:24 发布

原创最新推荐文章于 2019-09-18 20:15:24 发布 · 197 阅读

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本文介绍了一个简单的二维动态规划算法，用于判断字符串s3是否由s1和s2交织而成。通过构建二维布尔矩阵来逐步验证s3是否符合s1和s2的交织规则。

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

比较简单的二维DP，状态方程见代码。

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function   
        if(s1.length()+s2.length()!=s3.length()) return false; //剪枝
        vector<vector<bool> > answer (s1.length()+1,(vector<bool>)(s2.length()+1));
        
        if(s1.length()==0&&s2.length()==0) return (s3.length()==0);
        if(s1.length()==0) return (s2==s3);
        if(s2.length()==0) return (s1==s3);
        
        answer[0][0] = true;
        
        for(int i=1;i<=s1.length();i++)
        {
            answer[i][0] = (s1[i-1]==s3[i-1])&&answer[i-1][0];
        }
        
        for(int i=1;i<=s2.length();i++)
        {
            answer[0][i] = (s2[i-1]==s3[i-1])&&answer[0][i-1];            
        }
        
        for(int i=1;i<=s1.length();i++)
        {
            for(int j=1;j<=s2.length();j++)
            {
                if(s1[i-1]==s2[j-1]&&s2[j-1]==s3[i+j-1])
                {
                    answer[i][j] = answer[i-1][j]||answer[i][j-1];
                }
                else if(s1[i-1]==s3[i+j-1])
                {
                    answer[i][j] = answer[i-1][j];
                }
                else if(s2[j-1]==s3[i+j-1])
                {
                    answer[i][j] = answer[i][j-1];
                }
                else
                {
                    answer[i][j] = false;
                }
            }
        }
        
        return answer[s1.length()][s2.length()];
    }
};